Calculate $C=\sin3\alpha\cos\alpha$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$. Calculate $$C=\sin3\alpha\cos\alpha$$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$.
My idea was to find $\sin\alpha$ and $\cos\alpha$. Then we have $\sin3\alpha=3\sin\alpha-4\sin^3\alpha$. So $$\tan2\alpha=2=\dfrac{2\tan\alpha}{1-\tan^2\alpha}\iff\tan^2\alpha+\tan\alpha-1=0$$ This equation has solutions $\left(\tan\alpha\right)_{1,2}=\dfrac{-1\pm\sqrt5}{2}$ but as $\alpha\in(0^\circ;45^\circ)\Rightarrow$ $\tan\alpha=\dfrac{\sqrt5-1}{2}$. Now $\sin\alpha=\dfrac{\sqrt5-1}{2}\cos\alpha$ and plugging into $\sin^2\alpha+\cos^2\alpha=1$ got me at $\cos^2\alpha=\dfrac{2}{5-\sqrt5}$
 A: Consider a right triangle with acute angle $2\alpha$ and side lengths
\begin{align}
\text{opposite} &= 2\\
\text{adjacent} &= 1\\
\text{hypotenuse} &= \sqrt{2^2 + 1^2} = \sqrt{5}
\end{align}
Hence,
\begin{align}
\tan(2\alpha) &= \frac{2}{1} = 2\\
\cos(2\alpha) &= \frac{1}{\sqrt{5}}\\
\sin(2\alpha) &= \frac{2}{\sqrt{5}}
\end{align}
We can now compute
\begin{align}
C = \sin(3\alpha)\cos(\alpha) &= \cos(\alpha)\cdot(3\sin(\alpha) - 4\sin^3(\alpha))\\
& = 3\sin(\alpha)\cos(\alpha) - 4\sin^2(\alpha)\sin(\alpha)\cos(\alpha)\\
& = 3\frac{\sin(2\alpha)}{2} - 4\sin^2(\alpha)\frac{\sin(2\alpha)}{2}\\
& = \frac{\sin(2\alpha)}{2}(3 - 4\sin^2(\alpha))\\
& = \frac{\sin(2\alpha)}{2}(2(1 - 2\sin^2(\alpha)) + 1)\\
& = \frac{\sin(2\alpha)}{2}(2\cos(2\alpha) + 1)\\
& = \frac{\left(\frac{2}{\sqrt{5}}\right)}{2}\left(2\left(\frac{1}{\sqrt{5}}\right) + 1\right)\\
& = \boxed{\frac{2 + \sqrt{5}}{5}}
\end{align}
A: Hint:
Writing $A$ for $\alpha$
$$2C=2\sin3A\cos A=\sin4A+\sin2A$$
Now   Weierstrass substitution to find $\sin4A$
$$\tan2A=2\implies\dfrac{\sin2A}2=\dfrac{\cos2A}1=\pm\sqrt{\dfrac1{1^2+2^2}}$$
As $0^\circ\le2A\le90^\circ, \sin2A\ge0$
Can you take it from here?
A: For this problem, I like your work, through the conclusion that
$$\tan(\alpha) = \frac{\sqrt{5} - 1}{2}. \tag1 $$
In my opinion, the simplest approach to complete the problem is to forgo any attempt at elegance, and simply use the following identities (one of which you have already referred to):

*

*$\displaystyle \tan^2(\alpha) + 1 = \frac{1}{\cos^2(\alpha)}.$


*$\displaystyle \sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha).$
What is being asked for is
$$\sin(3\alpha)\cos(\alpha). \tag2 $$
To me, the simplest approach is to manually calculate both $\sin(\alpha)$ and $\cos(\alpha)$, and then use these calculations to evaluate the expression in (2) above.

As you indicated, because of the stated domain for $(\alpha)$, you know that $\sin(\alpha)$ and $\cos(\alpha)$ are both non-negative.  Using (1) above,
$$\tan^2(\alpha) = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \implies $$
$$\sec^2(\alpha) = \tan^2(\alpha) + 1 = \frac{5 - \sqrt{5}}{2} \implies $$
$$\cos^2(\alpha) = \frac{2}{5 - \sqrt{5}} \implies \tag3 $$
$$\cos(\alpha) = \sqrt{\frac{2}{5 - \sqrt{5}}}. \tag4 $$
Using (3), you also have that
$$\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \frac{2}{5 - \sqrt{5}} = \frac{3 - \sqrt{5}}{5 - \sqrt{5}} \implies $$
$$\sin(\alpha) = \sqrt{\frac{3 - \sqrt{5}}{5 - \sqrt{5}}}. \tag5 $$

Now, (2), (4), (5), and the $\sin(3\alpha)$ identity can be used to complete the problem.
$$\sin(3\alpha)\cos(\alpha) = \left[3\sin(\alpha) - 4\sin^3(\alpha)\right] \cos(\alpha) $$
$$= \sin(\alpha)\cos(\alpha) \times \left[3 - 4\sin^2(\alpha)\right] $$
$$ = \sqrt{\frac{3 - \sqrt{5}}{5 - \sqrt{5}}} \times 
\sqrt{\frac{2}{5 - \sqrt{5}}} \times 
\left[ ~3 - \left(4 \times \frac{3 - \sqrt{5}}{5 - \sqrt{5}} ~\right) ~\right]. \tag 6 $$
In (6) above, you know that $\left(3 - \sqrt{5}\right) \times 2 = \left(6 - 2\sqrt{5}\right) = \left(\sqrt{5} - 1\right)^2.$
Therefore, the 1st two factors in (6) above simplify to
$\displaystyle \frac{\sqrt{5} - 1}{5 - \sqrt{5}}.$
The 3rd factor in (6) above may be re-expressed as
$\displaystyle \frac{\left(15 - 3\sqrt{5}\right) - \left(12 - 4\sqrt{5}\right)}{5 - \sqrt{5}} = \frac{3 + \sqrt{5}}{5 - \sqrt{5}}.$
Putting this all together, the final computation is
$$\frac{\sqrt{5} - 1}{5 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{5 - \sqrt{5}}  = \frac{2 + 2\sqrt{5}}{30 - 10\sqrt{5}} $$
$$= \frac{1 + \sqrt{5}}{15 - 5\sqrt{5}} \times \frac{15 + 5\sqrt{5}}{15 + 5\sqrt{5}} = \frac{40 + 20\sqrt{5}}{100} = \frac{2 + \sqrt{5}}{5}.$$
